Question

A projectile is shot from the ground at an angle of `pi/6` and a speed of `2 m/s`. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

Answer 1

`"distance" = 0.184` `"m"`

Explanation:

We're asked to find the distance a projectile is from its starting point when it reaches its maximum height, given its initial velocity.

When the projectile is at its maximum height, its instantaneous `y`-velocity `v_y` is zero. We can use the equation

`v_y = v_(0y) -g t`

to find the time `t` when this occurs.

First, let's find the components of the initial velocity, for reference:

`v_(0x) = v_0cosalpha_0 = (2color(white)(l)"m/s")cos(pi/6) = color(red)(1.73` `color(red)("m/s"`

`v_(0y) = v_0sinalpha_0 = (2color(white)(l)"m/s")sin(pi/6) = color(green)(1.00` `color(green)("m/s"`

Plugging in our known values into the equation (`v_y = 0` and `g = 9.81` `"m/s"^2`), we have

`0 = color(green)(1.00` `color(green)("m/s") - (9.81color(white)(l)"m/s"^2)t`

`t = 0.102` `"s"`

The `x`- and `y`- positions of at this time are given by

`Deltax = v_(0x)t`

`Deltay = v_(0y)t - 1/2g t^2`

Plugging in known values, we have

`Deltax = (color(red)(1.73)color(white)(l)color(red)("m/s"))(0.102color(white)(l)"s")`

`= 0.177` `"m"`

`Deltay = (color(green)(1.00)color(white)(l)color(green)("m/s"))(0.102color(white)(l)"s") - 1/2(9.81color(white)(l)"m/s"^2)(0.102color(white)(l)"s")^2`

`= 0.0510` `"m"`

The distance from the starting point is thus

`r = sqrt((0.177color(white)(l)"m")^2 + (0.0510color(white)(l)"m")^2)`

`= color(blue)(0.184` `color(blue)("m"`